Hi,
Suppose we have a polynomail of degree n:
p(x) = a0 + a1x + a2x^2 + ... + anx^n
where n is an odd number and "an" (the constant) not equal 0. Show that this polynomail has a real root. You can assume that polynomails are continuous.

So it is pretty intuitive that p(x) as x-> inf is positive and as x-> -inf is negative. I just don't know what facts about polynomials to get
a p(a) = negative number and a p(b) = positive number. Then I know you can use the intermediate value theorem.

Thanks

Oct 13th 2007, 04:19 PM

Jhevon

Quote:

Originally Posted by tbyou87

Hi,
Suppose we have a polynomail of degree n:
p(x) = a0 + a1x + a2x^2 + ... + anx^n
where n is an odd number and "an" (the constant) not equal 0. Show that this polynomail has a real root. You can assume that polynomails are continuous.

So it is pretty intuitive that p(x) as x-> inf is positive and as x-> -inf is negative. I just don't know what facts about polynomials to get
a p(a) = negative number and a p(b) = positive number. Then I know you can use the intermediate value theorem.

Thanks

first you can assume (without loss of generality) that . then take the limit as x goes to infinity (this will make the polynomial go to infinity as well), then take the limit as x goes to negative infinity (this causes the polynomial to go to negative infinity as well, since it is an odd polynomial). now you can apply the intermediate value theorem, since clearly .

Oct 13th 2007, 05:16 PM

ThePerfectHacker

Say without lose of generality. Define the following sequence . The claim is that . To see that you can write the first term but the second term since it means this limit goes to . Thus there is a so that . Similarly define . And show thus there is a so that . Meaning we can choose an interval large enough so that the polynomial changes from -1 to 1. Which completes the proof.